Lesson: Quadrilaterals

The only time the diagonals of a rhombus are equal length is when that rhombus is a square (a square is a very special kind of rhombus in which the angles are all 90 degrees).

In all other cases, the diagonals of a rhombus are not equal length. To see what I mean, check out the rhombus here: http://i1168.photobucket.com/albums/r500/GMATPrepNow/rhombus_zpssslp0yuu...

It's a very tricky question to explain without a diagram, but here goes....

Since the perimeter is 24 and the 2 longest sides are twice the length of the 2 smallest sides, the parallelogram has the dimensions 4-8-4-8

The 120-degree angle in one corner tells us that the adjacent corner has a 60-degree angle (which is a special angle).

So, if we make the side with length 8 the BASE, then we can see that the height of the parallelogram = 2√3

So, the area of the big parallelogram (PQRS) = (8)(2√3) = 16√3

The area of the small parallelogram (TQUV) is HALF the area of the big parallelogram (PQRS)

So, the area of the small parallelogram (TQUV) = 16√3/2 = 8√3

One side of the small parallelogram is twice the other side. So, let's let 2x be the length of its base, and let x = length of the slanted side.

Since the angle between them is 60 degrees (as with the large parallelogram) we can use special triangle properties to determine that (x√3)/2 is the height of the small parallelogram.

At this point, we can say...

Area of TQUV = (2x)(x√3)/2 = 8√3 [from earlier calculations]

Simplify: x²√3 = 8√3

Divide both sides by √3 to get: x² = 8

So x = √8 = 2√2

In other words, the shorter side of the small parallelogram (TQUV) has length 2√2, which means the longer side has length 4√2

So, the TOTAL perimeter = 2√2 + 4√2 + 2√2 + 4√2 = 12√2 ≈ 17

Link to question: http://gre.myprepclub.com/forum/qotd-9-the-length-of-a-leg-of-an-isoscel...

The answer is A.
If you show me your calculations, perhaps we can see what the issue is.

Yes, the sides have to be longer than zero :-)

You bet!
Here's my step-by-step solution: https://gre.myprepclub.com/forum/qotd-9-the-length-of-a-leg-of-an-isosce...

Cheers,
Brent

Hi okookiez,

The quick answer is that, since the walkway is 3 feet wide, we must add 3 to one side of our rectangular region and add 3 to the other side of our region.

I created a step-by-step solution that (I believe) answers your question: https://gre.myprepclub.com/forum/the-gure-above-represents-a-rectangular...

Does that help?

Cheers,
Brent

The diagonals of a rectangle can create four 45-45-90 triangles ONLY WHEN the rectangle is a SQUARE.
If the rectangle is NOT a square, then the diagonals will NOT create four 45-45-90 triangles.

Here's a graphic that shows this: https://imgur.com/1yZFqov

Cheers,
Brent

This question requires some answer choices, but we can answer it nonetheless.

Let x = the width of the playground
Let y = the length of the playground

Since the area of the playground is 2400, we can write: xy = 2400

The running track has length 140.
Since the running track is only on 3 sides, we can write: x + x + y = 140

ASIDE: We can also write x + y + y = 140, but that route will give us the same answers.

We now have two equations: xy = 2400 and 2x + y = 140

Solve the second equation for y to get: y = 140 - 2x

Now take the first equation, xy = 2400 and replace y with (140 - 2x).
We get: x(140 - 2x) = 2400
Expand to get: 140x - 2x² = 2400
Rearrange to get: 2x² - 140x + 2400 = 0
Divide both sides by 2 to get: x² - 70x + 1200 = 0
Factor to get: (x - 30)(x - 40) = 0
So, EITHER x = 30 OR x = 40

Let's find the corresponding y-values by using our equation y = 140 - 2x
If x = 30, then y = 140 - 2(30) = 80
So, one possibility is that the playground has dimensions 30 by 80

If x = 40, then y = 140 - 2(40) = 60
So, another possibility is that the playground has dimensions 40 by 60

So, the longest possible length of the wall is 80 meters.

Cheers,
Brent

Question link: https://gre.myprepclub.com/forum/what-is-the-perimeter-in-meters-of-a-re...
Why did you set the perimeter equal to the area?

The question tells us that one rectangle has the same AREA as some other rectangle, and we're suppose to find the perimeter of that first rectangle.

Here's my full solution: https://gre.myprepclub.com/forum/what-is-the-perimeter-in-meters-of-a-re...

Cheers,
Brent

Question link: https://gre.myprepclub.com/forum/in-the-xy-plane-one-of-the-vertices-of-...

Notice that, to travel from the vertex at (2, 2), to the CENTER of the square at (6, 6), we must travel 4 units UP and 4 units to the RIGHT.

If we repeat the same steps, we'll arrive at the other vertex.

Does that help?

Cheers,
Brent

Question link: https://gre.myprepclub.com/forum/a-project-requires-a-rectangular-sheet-...

We need the ratio of the sides of the BIG rectangle to equal the ratio of the sides of a SMALL rectangle.

In other words, we want:
(longest length of BIG rectangle)/(shortest length of BIG rectangle) = (longest length of SMALL rectangle)/(shortest length of SMALL rectangle)

See my diagram at https://gre.myprepclub.com/forum/a-project-requires-a-rectangular-sheet-...

BIG RECTANGLE:
longest length = x
shortest length = y
So, ratio = x/y

SMALL RECTANGLE:
longest length = y
shortest length = x/2
So, ratio = y/(x/2)

The resulting EQUATION is: x/y = y/(x/2)

Does that help?

Cheers,
Brent

Here's my full solution: https://gre.myprepclub.com/forum/a-flat-rectangular-flower-bed-with-an-a...

Cheers,
Brent

Question link: https://gre.myprepclub.com/forum/qotd-9-the-length-of-a-leg-of-an-isosce...

We're told that the triangle is an ISOSCELES right triangle.
This means the two legs have the same length (and the hypotenuse, which is the longest side, has a different length)

Does that help?

Cheers,
Brent

Question link: https://gre.myprepclub.com/forum/qotd-10-the-length-of-each-side-of-rect...

The question says "The length of EACH side of rectangle R is an integer..."
So, all 4 sides have integer lengths.

Cheers,
Brent

Question link: https://gre.myprepclub.com/forum/what-is-the-area-of-the-region-shown-ab...

Since all of the given angles are 90 degrees, we can conclude that opposite sides are parallel and equal length.

Cheers,
Brent

Here's my full solution: https://gre.myprepclub.com/forum/a-b-and-c-are-three-rectangles-the-leng...

Cheers,
Brent

Here's my full solution: https://gre.myprepclub.com/forum/gre-math-challenge-794.html#p38357

Cheers,
Brent

Question link: https://gre.myprepclub.com/forum/rs-st-tu-ur-14558.html

Great question!

If we know the lengths of all sides in a square, then we CAN determine the length of the diagonal.
If we know the lengths of all sides in a rectangle, then we CAN determine the length of the diagonal.

If we know the lengths of all sides in a rhombus or parallelogram, then we CANNOT determine the length of the diagonal.

Cheers,
Brent

Hi Carla,

You don't need to know how to calculate area of a pentagon or hexagon.
As you suggest, if you're given such a question, you'll be able to divide the shape into easier shapes and then find those areas.

Cheers,
Brent

Hi Suryansh,

Sorry but I'm not sure what question you are referring to.
If you can provide a link to the question, I'd be happy to answer your question

Cheers,
Brent

In case 1, Quantity A is great than Quantity B (since 225 > 200).
Aside: at this point, the correct answer must be either A or D
In case 2, Quantity B is great than Quantity A (since 125 < 200).

D) The relationship cannot be determined from the information given.
Since we cannot definitively say which Quantity is great, the correct answer must be D.

Question link: https://gre.myprepclub.com/forum/a-at-rectangular-picture-represented-by...

The question asks us to find a value of x so that the area of the frame equals the area of the picture.
So: (frame area) = (picture area)

We also know that: (frame area) + (picture area) = area of entire rectangle
Since (frame area) = (picture area), we can also write: (picture area) + (picture area) = area of entire rectangle
In other words, 2(picture area) = area of entire rectangle

Does that help?

Cheers,
Brent